Find the area of the region in the first quadrant enclosed by the x-axis, the line y=x and circle x²+y²=32.

Given y=x — (1)
x²+y²=32 —(2)
Solving (1) and (2), we find that the line and the circle meet at B(4, 4) in the first quadrant as shown in the figure.
Draw perpendicular BM to the x-axis.
Therefore, the required area = area of the region OBMO + area of the region BMAB.
Now, the area of the region OBMO = ∫₄₀ ydx = ∫₄₀ xdx = ½[x²]₄₀ = 8 — (3)
Again, the area of the region BMAB, = ∫₄⁰ √(32-x²)dx = [½x√(32-x²) + ½ × 32 × sin⁻¹(x/4√2)]₄⁰
= [½(4)√(32-16) + ½ × 32 × sin⁻¹(1)] - [0 + ½ × 32 × sin⁻¹(0)]
= 8 + 16(π/2) - 0 = 8 + 8π — (4)
Adding (3) and (4), we get the required area = 8 + 8π - 8 = 8π sq.units.